In the disintegration of a radioactive element, `alpha`- and `beta`-particles are evolved from the nucleus.
`._(0)n^(1) rarr ._(1)H^(1) + ._(-1)e^(0) +` Antineutrino + Energy
`4 ._(1)H^(1) rarr ._(2)He^(4) + 2 ._(+1)e^(0) +` Energy
Then, emission of these particles changes the nuclear configuration and results into a daughter nuclide. Emission of an `alpha`-particles results into a daughter element having atomic number lowered by 2 and mass number by 4, on the other hand, emission of a `beta`-particle yields an element having atomic number raised by 1.
Which of the following combinations give finally an isotope of the parent element?

Options

A. `alpha, alpha, beta`
B. `alpha, gamma, alpha`
C. `alpha, beta, beta`
D. `beta, gamma, alpha`

Explanation

To solve the problem, we need to analyze the combinations of particle emissions from a radioactive element and determine which combination results in an isotope of the parent element. An isotope has the same atomic number but a different mass number.

### Step-by-Step Solution:

1. **Understanding Isotopes**:
- Isotopes are variants of a chemical element that have the same number of protons (atomic number) but different numbers of neutrons (mass number). Therefore, for an isotope of the parent element, the atomic number must remain the same while the mass number changes.

2. **Initial Element Representation**:
- Let the parent element be represented as \( X \) with atomic number \( A \) and mass number \( M \).

3. **Analyzing the Combinations**:
- We will evaluate each combination of particle emissions to see if we can return to an isotope of the parent element.

4. **Combination 1: Alpha, Alpha, Beta (α, α, β)**:
- **First Alpha Emission**:
- New Element: \( Y \) with atomic number \( A - 2 \) and mass number \( M - 4 \).
- **Second Alpha Emission**:
- New Element: \( Z \) with atomic number \( A - 4 \) and mass number \( M - 8 \).
- **Beta Emission**:
- New Element: \( W \) with atomic number \( A - 3 \) and mass number \( M - 8 \).
- **Conclusion**: Not an isotope (atomic number has changed).

5. **Combination 2: Alpha, Gamma, Alpha (α, γ, α)**:
- **First Alpha Emission**:
- New Element: \( Y \) with atomic number \( A - 2 \) and mass number \( M - 4 \).
- **Gamma Emission**:
- New Element: \( Y \) remains the same (no change).
- **Second Alpha Emission**:
- New Element: \( Z \) with atomic number \( A - 4 \) and mass number \( M - 8 \).
- **Conclusion**: Not an isotope (atomic number has changed).

6. **Combination 3: Alpha, Beta, Beta (α, β, β)**:
- **First Alpha Emission**:
- New Element: \( Y \) with atomic number \( A - 2 \) and mass number \( M - 4 \).
- **First Beta Emission**:
- New Element: \( Z \) with atomic number \( A - 1 \) and mass number \( M - 4 \).
- **Second Beta Emission**:
- New Element: \( X \) with atomic number \( A \) and mass number \( M - 4 \).
- **Conclusion**: This is an isotope of the parent element (same atomic number \( A \), different mass number \( M - 4 \)).

7. **Combination 4: Beta, Gamma, Alpha (β, γ, α)**:
- **First Beta Emission**:
- New Element: \( Y \) with atomic number \( A + 1 \) and mass number \( M \).
- **Gamma Emission**:
- New Element: \( Y \) remains the same (no change).
- **Alpha Emission**:
- New Element: \( Z \) with atomic number \( A - 1 \) and mass number \( M - 4 \).
- **Conclusion**: Not an isotope (atomic number has changed).

### Final Answer:
The combination that results in an isotope of the parent element is **Alpha, Beta, Beta (α, β, β)**.